640
840
520
920
Correct answer is B
Number of itineraries = 7P4
=\(\frac{7!}{(7-4)!}\\
=\frac{7!}{3!}\\
=\frac{7 \times 6 \times 5 \times 4 \times 3!}{3!}\\
=840\)
6/30
5/30
17/30
11/30
Correct answer is C
Total fruits 6 + 11 + 13 = 30,
prob. (Grape) = \(\frac{6}{30}\)
prob. (Banana) = \(\frac{11}{30}\)
= \(\frac{11}{30}\) + \(\frac{6}{30}\) = \(\frac{17}{30}\)
36
48
30
40
Correct answer is A
No explanation has been provided for this answer.
How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1-28?
12
10
9
8
Correct answer is C
3, -6, +12, -24
a = 3, r = -2
\(8n = \frac{a(1-r^n)}{1-r}\\
∴1-2^8 = \frac{3(1-(-2^{n-1}))}{1-(-2)}\\
1-2^8 = \frac{3(1-(-2^{n-1}))}{3}\)
1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9
If y = x\(^2\) - x - 12, find the range of values of x for which y \( \geq \) 0
x < -3 0r x > 4
x \( \leq \) -3 or x \( \geq \) 4
-3 < x \( \geq \) 4
-3 \( \leq \) x \( \leq \) 4
Correct answer is B
y = x\(^2\) - x - 12
= (x - 4)(x + 3)
∴ x = 4 or x = -3
Checking the cases for y \( \geq \) 0
We check values on the range x - 4 \(\geq\) 0; x + 3 \(\leq\) 0; x - 4 \(\leq\) 0 and x + 3 \(\geq\) 0 for the range which satisfies the inequality x\(^2\) - x - 12 \(\geq\) 0.
We find that the inequality is satisfied on the range x \(\leq\) -3 and x \(\geq\) 4.